This section is from the "Architectural Iron And Steel, And Its Application In The Construction Of Buildings" book, by WM. H. Birkmire.. Also see Amazon: Architectural Iron And Steel, And Its Application In The Construction Of Buildings.

There are several important points to be considered in planning a roof for a building. The trusses should be so placed as to be central over piers; consequently the number of trusses will be governed by the number of piers and the distance between centres, which should be from eight to sixteen feet; twelve feet being a suitable proportionate distance. The closer the trusses the less the proportion of the load each will have to sustain, and the lighter may be the rafters to support the roof surface.

In all calculations for roofs, the exact load should be found which will come upon each truss - the weight of the truss itself, the roof, ceiling, snow and wind. The rafters will be required to support the roofing timbers, the boarding, covering, snow and wind. The tie beam will have to support the weight of ceiling.

According to Mr. Trautwine, for spans not exceeding seventy-five feet, and with trusses seven feet apart, the total dead load per square foot, including the truss itself, purlins, etc., complete, may be safely taken as follows:

Pounds. | |

Roof covered with corrugated iron, unboarded.... | 8 |

If plastered below rafters................. | , 18 |

Roof covered with corrugated iron on boards...... | 11 |

If plastered below rafters..... | 18 |

35 |

Pounds. | |

Roof covered with slate, unboarded as on laths.... | 13 |

" " " " on boards 1 1/4 inches thick.... | 16 |

If plastered below rafters...... | 26 |

Roof covered with shingles or laths..... | 10 |

If plastered below the rafters or tie beam.... | 20 |

Roof covered with shingles on 1/8 inch boards...... | 13 |

For spans of from seventy-five to one hundred and fifty feet it will suffice to add four pounds to each of these totals.

For wind pressure taken at an angle of 22 1/2 degrees add 10 pounds per square foot; at an angle of 45 degrees, 20 pounds. For snow reverse these figures, adding for pressure at 22 1/2 degrees 20 pounds, and at 45 degrees 10 pounds per square foot.

If a ceiling composed of wooden timbers with lathing and plastering is to be included, 10 to 12 pounds per square foot should be added. If the ceiling rafters are to be of iron, their size and thickness may readily be calculated. Then take 7 to 9 pounds per square foot for plastering, II pounds for two-inch porous terra-cotta blocks: if metallic or wire lath is used instead of porous blocks, only slight allowance need be made therefor.

**The Graphic Method** consists of representing loads and strains by lines drawn to a scale of pounds to the fraction of an inch, being the simplest and readiest way of computing strains in trusses. Only a few examples will be given, a treatise on roofs not being within the scope of this volume.

No. 1 is a simple king-post truss with braces for supporting centre of the rafters. For finding the strains it is best to represent each member by a single line, being a diagram of the truss. Having found the correct weight to be supported, it should be divided into four parts; the point between D and F taking one half of F and D, the point between B and D one half of each, and each abutment one half of B and one half of H. The fine lines represent tension, the heavy lines compression.

With any convenient scale, as Fig. ia, lay off the spaces representing the vertical load supported at each point of the roof respectively.

As the reactions will be each one half the load, we locate k half way between the abutments.

Begin with the joint at foot of rafter and draw ab parallel to AB, and ak parallel to AK until it meets ab at a; draw cd parallel to CD; draw ac parallel to A C, and ce parallel to CE.

The joints to the right will be similar to those on the left, the truss being uniformly loaded.

The strains may be then scaled off, on which the total load = L as follows:

The length of | ab | ak | cd | ac | ce |

= strain on | AB | AK | CD | AC | CE |

(Fig. 2) Draw a strain diagram, Fig. 2A, as before.

Draw ab parallel to AB, ag parallel to AG, cd parallel to CD, ac parallel to A C, ce parallel to CE, eg parallel to EG.

The strains may be then scalfed off, on which the total load = L as follows:

The length of | ab | ag | cd | ac | ce | eg |

= strain on | AB | AG | CD | AC | CE | EG |

By continuing the strain diagram it will be noticed that the other side is similar; it is therefore unnecessary to continue the other half.

**Truss No. 3**(Fig. 1) is divided into eight parts, continuing the strain sheet as before, but using half of diagram Fig. ia, which would give bo. Draw ab parallel to AB, ao to AO, cd to CD, ac to AC, ck to CK; then draw ef parallel to EF, and gh to GH; then draw eg in line with ac, and draw kl parallel to KL, and el to EL. Then draw gm parallel to GM; it will pass through l. Draw om intersecting gm at m. Im lies over gm and is taken as two lines. MM is a light rod preventing the main tie from sagging. The strain may then be scaled off, on which the total load = L as follows:

The length of | ab | ao | cd | ac | ck | ef | gh | kl | el | gm | om | Im |

= strain on | AB | AO | CD | AC | CK | EF | GH | KL | EL | GM | OM | LM |

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